diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md
index 4bcbd1f9..dda7f2ad 100644
--- a/problems/面试题02.07.链表相交.md
+++ b/problems/面试题02.07.链表相交.md
@@ -214,7 +214,41 @@ class Solution:
         return head
 ```
 ```python
-（版本三）等比例法
+（版本三）求长度，同时出发 （代码复用 + 精简）
+class Solution:
+    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
+        dis = self.getLength(headA) - self.getLength(headB)
+        
+        # 通过移动较长的链表，使两链表长度相等
+        if dis > 0:
+            headA = self.moveForward(headA, dis)
+        else:
+            headB = self.moveForward(headB, abs(dis))
+        
+        # 将两个头向前移动，直到它们相交
+        while headA and headB:
+            if headA == headB:
+                return headA
+            headA = headA.next
+            headB = headB.next
+        
+        return None
+    
+    def getLength(self, head: ListNode) -> int:
+        length = 0
+        while head:
+            length += 1
+            head = head.next
+        return length
+    
+    def moveForward(self, head: ListNode, steps: int) -> ListNode:
+        while steps > 0:
+            head = head.next
+            steps -= 1
+        return head
+```
+```python
+（版本四）等比例法
 # Definition for singly-linked list.
 # class ListNode:
 #     def __init__(self, x):